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Understanding Work Done in Physical Systems

Explore the concept of work in physical systems through practical examples and theoretical explanations to deepen your understanding. Learn how forces, distances, and angles impact work done and energy transfer. Enhance your knowledge of the Work-Kinetic Energy Theorem and practical applications in stopping a car.

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Understanding Work Done in Physical Systems

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  1. Chapter 7Energy of a System EXAMPLES

  2. Example 7.1 Conceptual Example • If the magnitude of F is held constant but the angle θ is increased, What happened with the work W done by F? DECREASES!!! • Since: cosθDecreases when 0 < θ < 90o Δr

  3. Example 7.2 Conceptual Example • Can exert a force & do no work! • Example: Walking at constant v with a grocery bag. W = FΔr cosθ • Could have Δr = 0, F ≠ 0  W = 0 Could have F  Δr  θ= 90º, cosθ = 0  W = 0 Δr

  4. Example 7.3 Work Done ON a Box If: m = 50 kg, FP = 100N, Ffr = 50N, Δr = 40.0m Find: (a).Work done BY each force. (b). Net work Done ON the box. • For each force ON the box: W = F Δr cosθ (a) WG = mgΔrcos90o = mg(40m)(0) = 0 Wn = nΔrcos90o = n(40m)(0) = 0 WP = FPΔrcos37o = (100N)(40m)(cos37o) = 3200J Wfr= FfrΔrcos180o = (50N(40m)(–1)= = – 2000J (b) • Wnet= WG+Wn + WP+ Wfr= 1200J • Wnet=(Fnet)xΔr = (FPcos37o – Ffr) Δr  Wnet= (100Ncos37o – 50N)40m= 1200J

  5. Example 7.4 Work Done ON a Backpack • Find The Net Work done ONthe backpack. • From (a): h = dcosθ • From (b): ΣFy = 0  FH = mg • From (c): • Work doneBY the hiker: WH=FHdcosθWH=mgh • From (c): • Work doneBY gravity: WG = mg dcos(180 – θ) WG = mg d(–cosθ) = – mg dcosθ WG =–mgh • NET WORK on the backpack: Wnet = WH+ WG = mgh – mgh Wnet= 0

  6. Example 7.5 Work BY the Earth ON the Moon • FG exerted BY the Earth ON the Moon acts toward the Earth and provides its centripetal acceleration inward along the radius orbit • FG  Δr (Tangent to the circle & parallel with v) The angle θ = 90o WE-M = FGΔr cos 90o = 0 • This is why the Moon, as well as artificial satellites, can stay in orbit without expenditure of FUEL!!!

  7. Example 7.6 Work Done by a Constant Force (Example 7.3 Text Book) Given: Dr = (2.0 î +3.0 ĵ) m F = (5.0 î +2.0 ĵ) N • Calculate the following magnitudes: • Δr= (4 + 9)½ = (13)½ = 3.6 m • F= (25 + 4)½ = (29)½ = 5.4 N • Calculate the Work done by F: W = F•Δr = [(5.0 î +2.0 ĵ) N][(2.0 î +3.0 ĵ) m] = (5.0 î •2.0 î + 5.0 î •3.0 ĵ + 2.0 ĵ •2.0 î + 2.0 ĵ •3.0 ĵ) N •m = [10 + 0 + 0 + 6] J =16 J

  8. Example 7.7 Net Work Done from a Graph (Example 7.4 Text Book) • The Net Work done by this force is the area under the curve W = Area under the Curve W = AR + AT W =(B)(h)+(B)(h)/2 =(4m)(5N)+(2m)(5N)/2 W = 20J + 5J = 25 J

  9. Example 7.8 Work-Kinetic Energy Theorem (Example 7.6 Text Book) • m = 6.0kg first at rest is pulled to the right with a force F = 12N (frictionless). • Find v after m moves 3.0m • Solution: • The normal and gravitational forces do no work since they are perpendicular to the direction of the displacement • W = F Δx =(12)(3)J = 36J W = ΔK = ½ mvf2 – 0 36J = ½(6.0kg)vf2 = (3kg)vf2 Vf =(36J/3kg)½ = 3.5m/s

  10. Example 7.9 Work to Stop a Car • Wnet = Fdcos180°= –Fd= –Fd Wnet = K= ½mv22 – ½mv12 = –Fd  – Fd = 0 – ½m v12  dv12 • If the car’s initial speed doubled, the stopping distance is 4 times greater. Then: d = 80 m

  11. Material for the Final Exam • Examples to Read!!! • Example 7.5 (page 175) • Example 7.7 (page 179) • Material from the book to Study!!! • Objective Questions: 7-11-14 • Conceptual Questions: 2-5-8 • Problems: 1-6-9-14-15-21-30-33-42-44

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